## Precalculus (6th Edition) Blitzer

$3\sqrt{3}$
RECALL: For any non-negative real numbers $a$ and $b$, $\sqrt{ab} = \sqrt{a}\cdot \sqrt{b}$ and $\sqrt{a} \cdot \sqrt{b} = \sqrt{ab}$. Simplify the given expression by using the rule above to obtain $\sqrt{9} \cdot \sqrt{3} \\=\sqrt{3^2} \cdot \sqrt{3} \\=3\sqrt{3}.$